How do you find the integral of #dx/ sqrt(x^2 - a^2)#?

Question

Luke Alvarez · Accepted Answer

With this general case, notice how #sqrt(x^2 - a^2) prop sqrt(sec^2theta - 1)#. So, we can use the following substitution: #x = asectheta#
#dx = asecthetatanthetad theta#
#sqrt(x^2 - a^2) = sqrt(a^2sec^2theta - a^2) = atantheta# Thus, we have: #= int 1/(cancel(atantheta))*cancel(a)secthetacancel(tantheta)d theta# #= int secthetad theta# Then just a little trick:
#= int sectheta((sectheta + tantheta)/(sectheta + tantheta))d theta# #= int (sec^2theta + secthetatantheta)/(sectheta + tantheta)d theta# Now, let:
#u = sectheta + tantheta#
#du = secthetatantheta + sec^2thetad theta# Therefore: #= int1/udu# #= ln|u|# #= ln|sectheta + tantheta|# Recall that:
#sectheta = x/a#
#tantheta = sqrt(x^2 - a^2)/a# Thus we have: #= color(green)(ln|x/a + sqrt(x^2 - a^2)/a| + C)# ...which is perfectly acceptable. But, you could simplify this more and be a little sneaky. #= ln|(1/a)[x + sqrt(x^2 - a^2)]| + C# #= ln|x + sqrt(x^2 - a^2)| + ln|1/a| + C# but since #a# is a constant... it gets embedded into #C#. #= color(blue)(ln|x + sqrt(x^2 - a^2)| + C)# So if you see Wolfram Alpha give you this answer, that's why.

Axel Alvarez · Answer

undefined To find the integral of $ \frac{dx}{\sqrt{x^2 - a^2}} $, where $ a $ is a constant, you can use a trigonometric substitution. Let $ x = a \sec(\theta) $, then $ dx = a \sec(\theta) \tan(\theta) d\theta $. Substituting these into the integral, you get $ \int \frac{a \sec(\theta) \tan(\theta) d\theta}{\sqrt{a^2 \sec^2(\theta) - a^2}} $. Simplifying the expression under the square root gives $ \int \frac{a \sec(\theta) \tan(\theta) d\theta}{\sqrt{a^2(\sec^2(\theta) - 1)}} $, which becomes $ \int \frac{a \sec(\theta) \tan(\theta) d\theta}{\sqrt{a^2 \tan^2(\theta)}} $. This simplifies further to $ \int d\theta $. After integrating with respect to $ \theta $, you can convert back to the original variable $ x $.